sn2/e2 q

[gomawum]

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I'm doing this 3 sets of full test my brother got for me and it has so many wrong answers which drives me nuts.
and to make it worst, it added me a far more confusion on the stuffs that I was sure about... and now i'm not....

so I decided to throw it in the garbage after I clarify few couple concepts that I became unclear after this...

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View attachment 9991

(the cross means + )
for this rxn, if the solventis T-BuOH,

does it make it sn2? or E2?

I know leaving group is 1' and base is also 1' so the rxn is sn2
but the T-BuOH make me feel like the rxn is E2..
(the protic solvent for sn2 is not favorable anywayz..)

if you have to choose one, which one is more likely it?

(I'm tempting to answer the rxn is elimination because of steric hindrance of solvent, so if it is, does the all the sn2 rxn with primary leaving group with primary base but hindered solvent makes it elimination always? can someone clarify it?)

 

[Boat on the Chesapeake]

The attachment doesn't seem to work, but I think I agree with your reasoning that it's E2.

 

[bigstix808]

your reasoning makes sence for pushing a E2 rxn, but without knowing what the base is reacting in the rxn (attachement doesn't work) it is really hard to say. but given the information we have, i'm reluctant to push off this problem as being 100% E2 because 1) if weak base then SN2 will be prefered, 2) 1° reactant will prefer SN2 because steric henderance is not a factor in E2 reactions and elimination is more likely to occur on a 3° reactant not a 1° as you have stated is present here. you are correct in assuming that SN2 prefers to have a polar aprotic solvent, but that does not mean the rxn will not take place when given a polar protic one (such as t-butanol). and one last note, when SN2 and E2 are in competition, steric bulk in either the base or nucleophile (which we don't know - unless t-butanol is our nucleophile and not the solvent) will prevent SN2 and favor E2.

hope that helps...

if you can get that attachment working i'm sure you can get a bit more advice on your dilemma.

 

[gomawum]

it's a 3-bromo-propyl-cyclohexane with -OCH2CH3 ---->(solvent is T-BuOH)

so the leaving group is 1' and base is strong and not steric-hindered, but it's the T-BuOH solvent that is bugging me..
strong base with 1' leaving group usually go to SN2,
but what is that T-BuOH(polar protic which is not favored in either sn2/e2) has to do with the rxn?

 

[gomawum]

and also..

when is the case that elimination reaction is the anti-markovnikov?

someone told me
tertiary leaving group with tertiary base(T-BuOH or T-BuO-) and
2ndary leaving group with tertiary base (T-BuOH or T-BuO-) in both E2 and E1 alwayls make anti...

is it true?

 

[klutzy1987]

and also..

when is the case that elimination reaction is the anti-markovnikov?

someone told me
tertiary leaving group with tertiary base(T-BuOH or T-BuO-) and
2ndary leaving group with tertiary base (T-BuOH or T-BuO-) in both E2 and E1 alwayls make anti...

is it true?

Because t-botoxide is big and bulky it cuts to the outside and therefor forms the hoffman. A smaller cutting tool like methoxide cuts to the inside and forms the more stable alkene, The Zaitchev.